将时间存储在DB中并在登录PHP后恢复

Question

We already know that the following code in PHP will log the user out after 5 mins of inactivity.

$timeout = 5*60; // Set timeout minutes
$logout_redirect_url = "index.php"; // Set logout URL

if (isset($_SESSION['start_time'])) {
    $elapsed_time = time() - $_SESSION['start_time'];
    if ($elapsed_time >= $timeout) {
        session_unset();
        session_destroy();
        header("Location: $logout_redirect_url");
    }
}
$_SESSION['start_time'] = time();

I want to implement a modification of the current code and do something like this:

• Assume the user logs out when he had 3 minutes left before automatic logout(assuming the time doesn't restart for him after his inactivity for 2 minutes), we keep track of the time he has left by storing it in a DB (MySQL) and later on start reducing from the same 3 minutes after he logs back in. How can i do this?

Answer 1

Track by the time used, not the currentTime/storedTime. Just use those to figure out the time remaining. This is a quick example. There may be some small errors and improvements that can be made. It should be plenty to help you implement a solution.

User visits page:

if (empty($_SESSION['start_time'])) {
  $_SESSION['start-time'] = time();
}

$timeLeft = //get time from db

//if there is a value in the db, that is the time left, otherwise, use the max time allowed (new timer)
$timeLeft = (!empty($timeLeft)) ? $timeLeft : $timeAllowed
$timePassed = time() - $_SESSION['start_time'];
if ($timePassed > $timeAllowed) {
  //logout
}

Then, when the user leaves:

$timeLeft = $timeAllowed - (time() - $_SESSION['start_time']);
//Store $timeLeft in the database - should be a value like 180 (3 minutes)